The Particle in a Three-Dimensional Box

The infinite square well is the first exactly solvable problem in quantum mechanics, but it lives in one dimension. Real systems — atoms in a trap, electrons in a metal, a gas in a container — occupy three. The particle in a three-dimensional box is the cleanest bridge: a particle confined to a rectangular region with impenetrable walls, free inside and forbidden outside. Because the walls act independently along each Cartesian axis, the Schrödinger equation separates into three one-dimensional problems, and the full solution is a product of the wavefunctions you already know. The payoff is not the energies alone but their pattern. In a cubic box the energy depends only on the sum $n_1^2+n_2^2+n_3^2$ , so many distinct states — distinguished by which direction carries the excitation — share one energy. This is degeneracy, and it is the central new phenomenon of the three-dimensional problem. Most of it is forced by the cube's symmetry: the three axes are interchangeable, so permuting the quantum numbers cannot change the energy. But not all of it. At the level $n_1^2+n_2^2+n_3^2=27$ the states $(1,1,5)$ and $(3,3,3)$ collide even though neither is a permutation of the other — a genuine accidental degeneracy that symmetry does not explain. This article builds the solution by separation of variables, counts the degeneracies from the ground state up, exhibits the accidental collision, and then lets the box grow large. In that limit the discrete levels crowd together into a smooth density of states that scales as the volume — the very quantity that, filled with electrons up to a Fermi surface, becomes the free-electron model of a metal. The three-dimensional box is a warm-up that already contains the seeds of solid-state physics.

Key takeaways

A particle in a rigid rectangular box separates: $\Psi(x,y,z)=X(x)\,Y(y)\,Z(z)$ , each factor a one-dimensional infinite-well eigenfunction, and the energy is the sum of the three 1D energies.
For a cubic box of side $L$ , $E_{n_1n_2n_3}=\dfrac{\hbar^2\pi^2}{2mL^2}\,(n_1^2+n_2^2+n_3^2)$ with each $n_i=1,2,3,\dots$ ; the ground state $(1,1,1)$ has energy $3E_0$ where $E_0=\hbar^2\pi^2/2mL^2$ .
The energy depends only on $N=n_1^2+n_2^2+n_3^2$ , so states that permute the quantum numbers are degenerate: $g=1,3,3,3,1,6,\dots$ for the lowest levels.
Most degeneracy comes from the cube's symmetry, but accidental degeneracy also occurs — the clearest case is $(1,1,5)$ and $(3,3,3)$ , both giving $N=27$ though one is not a permutation of the other.
Letting the box grow, the level count below energy $E$ scales as the volume, giving a density of states $\propto V\sqrt{E}$ — the entry point to the free-electron model of metals.
Key idea: confinement in each Cartesian direction is independent, so a 3D box is three 1D boxes multiplied together — and the interchangeability of the axes is exactly what produces the rich degeneracy structure.

The core insight

The three-dimensional box separates into three independent one-dimensional infinite wells; the total energy is the sum of the three, and because that sum is symmetric under permuting the axes, distinct states share energies — the origin of quantum degeneracy.

Where this lives today

The three-dimensional box is a workhorse well beyond its textbook role. Its density of states — the number of levels per unit energy, growing as $V\sqrt{E}$ — is the starting point of the free-electron model of metals: fill the box states with electrons up to the Fermi energy and you predict electronic heat capacity, Pauli paramagnetism, and the Fermi surface, all before adding any lattice potential. The same counting underlies blackbody radiation (photon modes in a cavity), the Debye model of phonons (vibrational modes in a crystal), and the Sackur–Tetrode entropy of an ideal gas. In nanoscience the box is nearly literal: a quantum dot is an electron confined to a few-nanometer region, and its discrete, size-tunable levels — exactly the $E\propto(n_1^2+n_2^2+n_3^2)/L^2$ scaling derived here — are engineered to emit specific colors in displays and biological markers. Degeneracy, the article's central theme, reappears everywhere a system has symmetry, and the accidental degeneracy first met at $N=27$ foreshadows the hidden symmetries that make problems like the hydrogen atom and the isotropic oscillator exactly solvable.

The physical picture

A particle sits inside a rectangular region $0<x<L_1$ , $0<y<L_2$ , $0<z<L_3$ , with the potential zero inside and infinite outside. Infinite walls force the wavefunction to vanish on every face, exactly as in one dimension. Inside, the particle is free, so the time-independent Schrödinger equation is $-\frac{\hbar^2}{2m}\nabla^2\psi=E\psi$ with $\nabla^2=\partial_x^2+\partial_y^2+\partial_z^2$ . The crucial structural fact is that this operator is a sum of pieces, one per axis, and the boundary conditions decouple axis-by-axis. That is precisely the situation in which separation of variables works: a product ansatz turns one three-dimensional partial differential equation into three one-dimensional ordinary ones. Each factor is a familiar standing wave — $X(x)\propto\sin(n_1\pi x/L_1)$ and likewise for $y,z$ — vanishing at both walls and carrying an integer number of half-wavelengths. The energy is additive because the kinetic energy is: the particle's motion along $x$ , $y$ , $z$ contributes independently. The whole novelty of three dimensions is therefore not the individual factors but how their energies combine, and what happens when the box is symmetric enough that different combinations tie.

Figure

A 3D box separates into three independent 1D wells

The degeneracy structure is easiest to see as a lattice. Each state is a point $(n_1,n_2,n_3)$ with positive-integer coordinates, and its energy is proportional to the squared distance from the origin, $N=n_1^2+n_2^2+n_3^2$ . States of equal energy therefore lie on a sphere of fixed radius $\sqrt N$ . Counting degeneracy becomes counting lattice points on a sphere — a number-theoretic question, which is exactly why "accidental" coincidences appear: two lattice points at the same radius need not be related by any symmetry of the cube.

Figure

Degeneracy as lattice points on a sphere of radius √N

The core equations

Start from the free Schrödinger equation inside the box, where the potential vanishes and only the Laplacian appears. This is the equation whose separability drives everything that follows.

-\frac{\hbar^2}{2m}\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right)\psi = E\,\psi \qquad (1)

where: $\hbar$ is the reduced Planck constant; $m$ is the particle mass; $\psi(x,y,z)$ is the stationary-state wavefunction; $E$ is the energy eigenvalue; and the parenthesis is the three-dimensional Laplacian $\nabla^2$ . Outside the box $\psi\equiv0$ , and continuity forces $\psi$ to vanish on all six faces. Physically, the particle is free inside but perfectly reflected at every wall. Separation of variables produces normalized product eigenfunctions, one standing wave per axis, each vanishing at its two walls.

\psi_{n_1n_2n_3}(x,y,z)=\sqrt{\frac{8}{L_1L_2L_3}}\,\sin\!\frac{n_1\pi x}{L_1}\,\sin\!\frac{n_2\pi y}{L_2}\,\sin\!\frac{n_3\pi z}{L_3} \qquad (2)

where: $L_1,L_2,L_3$ are the box side lengths; $n_1,n_2,n_3=1,2,3,\dots$ are three independent positive integers; and the prefactor $\sqrt{8/L_1L_2L_3}$ normalizes the product to unit probability over the box. Each quantum number counts the half-wavelengths along one axis; setting any $n_i=0$ would kill the wavefunction, so the counting starts at one. The corresponding energies add, because the kinetic energy along each direction is independent.

E_{n_1n_2n_3}=\frac{\hbar^2\pi^2}{2m}\left(\frac{n_1^2}{L_1^2}+\frac{n_2^2}{L_2^2}+\frac{n_3^2}{L_3^2}\right) \qquad (3)

where: each term is the energy of the corresponding one-dimensional infinite well. The energy is a sum of squares, so the levels form a three-dimensional grid rather than a single ladder — the structural difference from the 1D problem. For the symmetric special case $L_1=L_2=L_3=L$ (the cubic box), the energy collapses to a single integer combination, and this is where degeneracy appears.

E_{n_1n_2n_3}=\frac{\hbar^2\pi^2}{2mL^2}\,\big(n_1^2+n_2^2+n_3^2\big)\equiv E_0\,N,\qquad N\equiv n_1^2+n_2^2+n_3^2 \qquad (4)

where: $E_0=\hbar^2\pi^2/2mL^2$ is the single-axis ground energy and $N$ is the integer "shell index." Two states with the same $N$ have the same energy regardless of how the excitation is distributed among the axes — the defining feature of the cubic box.

The core equations — the degeneracy count

The number of states at a given energy is the number of ordered triples of positive integers whose squares sum to $N$ ; we denote it $g(N)$ .

g(N)=\#\{(n_1,n_2,n_3)\in\mathbb{Z}_{>0}^3 : n_1^2+n_2^2+n_3^2=N\} \qquad (5)

where: the count is over ordered triples, so $(2,1,1)$ , $(1,2,1)$ , $(1,1,2)$ are three distinct states. The lowest values are $g(3)=1$ , $g(6)=g(9)=g(11)=3$ , $g(12)=1$ , $g(14)=6$ : the degeneracy jumps around erratically because it is a lattice-point count on a sphere, not a smooth function of energy.

Derivation ASeparation of variables solves the box exactly

The goal of this derivation is to show that the three-dimensional problem is not genuinely three-dimensional at all: the product ansatz $\psi=X(x)Y(y)Z(z)$ turns the single partial differential equation (1) into three independent one-dimensional eigenvalue problems, each identical to the ordinary infinite square well. The key move is to divide the full equation by the product $\psi$ so that it splits into three terms, each depending on only one variable; since a function of $x$ alone can equal a constant only if it is a constant, each term must separately be constant, and the total energy is their sum. We then import the known 1D well solution three times.

Step 1. Insert the product $\psi=X(x)Y(y)Z(z)$ into (1) and divide through by $XYZ$ . Because $\partial_x^2$ acts only on $X$ , and likewise for the others, the equation becomes $-\frac{\hbar^2}{2m}\big(\tfrac{X''}{X}+\tfrac{Y''}{Y}+\tfrac{Z''}{Z}\big)=E$ . Each ratio depends on a different variable.

Step 2. Argue term-by-term. The first term depends only on $x$ , the second only on $y$ , the third only on $z$ , yet they sum to the constant $E$ . Varying $x$ alone, while holding $y$ and $z$ fixed, cannot change the sum, so the $x$ -dependent term cannot vary with $x$ and so $-\frac{\hbar^2}{2m}\frac{X''}{X}=E_x$ is itself a constant; the same holds for $y$ and $z$ . Thus $E=E_x+E_y+E_z$ . Why this step is legal: a relation of the form $f(x)+g(y)+h(z)=\text{const}$ over an open region forces each of $f,g,h$ to be constant, since the partial derivatives $\partial_x,\partial_y,\partial_z$ of the constant all vanish.

Step 3. Solve each one-dimensional problem. The equation $X''=-k_1^2 X$ with $k_1^2=2mE_x/\hbar^2$ and boundary conditions $X(0)=X(L_1)=0$ is exactly the infinite square well: $X\propto\sin(n_1\pi x/L_1)$ with $k_1=n_1\pi/L_1$ and $E_x=\hbar^2\pi^2 n_1^2/2mL_1^2$ , $n_1=1,2,\dots$

Step 4. Assemble. Multiplying the three normalized factors gives (2); adding the three energies gives (3). The counting starts at $n_i=1$ because $n_i=0$ makes the corresponding sine vanish identically, which is not a physical state. The product ansatz indeed solves (1) with eigenvalue (3), and the lowest ordered-triple energies reproduce the level sequence $N=3,6,9,11,12,14$ used below.

Go deeper · Derivation B · 2 minSymmetry degeneracy versus accidental degeneracy

Derivation BSymmetry degeneracy versus accidental degeneracy

Here, the aim is to separate the degeneracy of the cubic box into two physically distinct kinds and to prove that the second kind genuinely exists. The first kind is forced by the cube's symmetry group: interchanging any two axes is a symmetry of the Hamiltonian, so permuting the quantum numbers must give a state of the same energy. The second kind — accidental degeneracy — is a coincidence of the number-theoretic function $N=n_1^2+n_2^2+n_3^2$ that no symmetry of the cube explains. The route is to first count the permutation multiplicities, then exhibit a level where two states of different sorted quantum numbers share $N$ , which no relabeling of axes can connect.

Step 1. Count symmetry degeneracy. For a triple with all three quantum numbers distinct, there are $3!=6$ orderings; for a triple with exactly two equal, there are $3$ orderings; for all three equal, there is $1$ . So $(1,2,3)$ -type levels have $g\ge6$ , $(2,2,1)$ -type have $g\ge3$ , and $(1,1,1)$ -type have $g=1$ from this source alone. This accounts for $g(3)=1$ , $g(6)=g(9)=g(11)=3$ , $g(12)=1$ , $g(14)=6$ .

Step 2. Look for a collision that permutations cannot produce. Compute $N$ for $(1,1,5)$ : $1+1+25=27$ . Compute $N$ for $(3,3,3)$ : $9+9+9=27$ . These are equal, but $\{1,1,5\}$ and $\{3,3,3\}$ are different multisets — no reordering of one gives the other.

Step 3. Conclude the total multiplicity at $N=27$ . The permutations of $(1,1,5)$ contribute $3$ states and $(3,3,3)$ contributes $1$ , so $g(27)=4$ , but this $4$ is not a single permutation orbit: it is two orbits that happen to sit at the same radius. That is accidental degeneracy — a lattice-point coincidence on the sphere $n_1^2+n_2^2+n_3^2=27$ . Why this matters: symmetry degeneracy is "explained" (a symmetry operator maps the states into each other), while accidental degeneracy signals either a hidden symmetry or, as here, a pure coincidence of sums of squares; distinguishing them is a recurring theme (the hydrogen atom's "accidental" $\ell$ -degeneracy turns out to hide the Runge–Lenz symmetry). *Enumerating all ordered triples with $N=27$ yields exactly the multiset $\{(1,1,5)\text{-perms}, (3,3,3)\}$ with $g(27)=4$ , confirming the two sorted triples are distinct.*

Limiting cases

Each limit recovers a simpler or complementary picture and checks the construction.

Case	Result	Meaning
Two dimensions squeezed, $L_2,L_3\to0$	Energy $\to\hbar^2\pi^2 n_1^2/2mL_1^2$	The transverse modes cost enormous energy and freeze out; the 1D infinite well is recovered.
Rectangular box, all $L_i$ different	Generic $g=1$ (accidental ties possible)	Breaking the cube's symmetry lifts the permutation degeneracy: distinct axes give distinct energies.
Cubic box	$g=1,3,3,3,1,6,\dots$	The three equal sides make the axes interchangeable, restoring the symmetry degeneracy.
Large box / high energy	Density of states $g(E)\,dE\propto V\sqrt{E}\,dE$	Discrete levels crowd into a smooth continuum; the count below $E$ scales with the volume $V=L^3$ .

Common trapDegeneracy is not just "the number of permutations"

It is tempting to compute a level's degeneracy by counting the orderings of one quantum-number triple — $6$ for all-distinct, $3$ for two-equal, $1$ for all-equal. That captures the symmetry degeneracy but misses accidental degeneracy, where two inequivalent triples land on the same energy. The cleanest example is $N=27$ : $(1,1,5)$ contributes $3$ states and $(3,3,3)$ contributes $1$ , for a total $g(27)=4$ that is not a single permutation orbit. Always count by solving $n_1^2+n_2^2+n_3^2=N$ over all triples, not by permuting one you happened to find. A second trap: the quantum numbers start at $1$ , never $0$ — a zero kills the wavefunction, so there is no " $n=0$ " ground state in a box.

Worked examples

Example 1 — Degeneracy of the third excited level of a cubic box. List the states and give the degeneracy of the third excited energy level.

Step 1. Order the levels by $N=n_1^2+n_2^2+n_3^2$ . The smallest achievable values are $N=3$ (ground, $(1,1,1)$ ), then $N=6$ , $N=9$ , then $N=11$ . So the ground state is $N=3$ , the first excited is $N=6$ , the second is $N=9$ , and the third excited level is $N=11$ .

Step 2. Enumerate all ordered triples with $n_1^2+n_2^2+n_3^2=11$ . The only unordered solution is $\{1,1,3\}$ since $1+1+9=11$ (no other combination of three positive squares gives $11$ ). Its orderings are $(1,1,3),(1,3,1),(3,1,1)$ .

Step 3. Count. Three ordered triples means degeneracy $g=3$ , with energy $E=11E_0=11\,\hbar^2\pi^2/2mL^2$ . The three states are physically distinct — they differ in which axis carries the doubly-excited mode — but share one energy. The verifier confirms $g(11)=3$ and the level ordering $N=3,6,9,11$ .

Example 2 — Ground-state energy of an electron in a $1\ \mathrm{nm}$ cubic box. Estimate the ground-state energy for an electron confined to a cube of side $L=1.0\ \mathrm{nm}$ .

Step 1. The single-axis unit is $E_0=\dfrac{\hbar^2\pi^2}{2m_eL^2}$ . With $\hbar=1.055\times10^{-34}\ \mathrm{J\,s}$ , $m_e=9.109\times10^{-31}\ \mathrm{kg}$ , $L=1.0\times10^{-9}\ \mathrm{m}$ : $E_0=\dfrac{(1.055\times10^{-34})^2\pi^2}{2(9.109\times10^{-31})(1.0\times10^{-9})^2}\ \mathrm{J}$ .

Step 2. Evaluate. The numerator is $(1.113\times10^{-68})(9.870)=1.099\times10^{-67}$ ; the denominator is $2(9.109\times10^{-31})(1.0\times10^{-18})=1.822\times10^{-48}$ . So $E_0=6.03\times10^{-20}\ \mathrm{J}=0.377\ \mathrm{eV}$ .

Step 3. The ground state $(1,1,1)$ has $N=3$ , so $E_{111}=3E_0=1.13\ \mathrm{eV}$ . A single confined electron therefore already carries over an electron-volt of zero-point energy — the physical reason quantum dots have optical gaps in the visible/near-infrared and why nanoscale confinement shifts colors. The takeaways are third excited level $N=11$ with $g=3$ , and a $1\ \mathrm{nm}$ box giving $E_0\approx0.38\ \mathrm{eV}$ , ground state $\approx1.1\ \mathrm{eV}$ .

Origins · 1 mindegeneracy and how this was discovered

degeneracy

structural concept · when distinct quantum states share one energy

Degeneracy is the phenomenon in which two or more physically distinct quantum states have exactly the same energy eigenvalue. It is the organizing concept of the three-dimensional box: the cube's interchangeable axes force states related by permuting the quantum numbers to be degenerate, and the number-theoretic structure of $n_1^2+n_2^2+n_3^2$ produces additional "accidental" coincidences. Degeneracy is far more than bookkeeping — the size of a degenerate multiplet sets statistical weights, its splitting under a perturbation is the whole subject of degenerate perturbation theory, and an "accidental" degeneracy is often the fingerprint of a hidden symmetry, as with the extra $\ell$ -degeneracy of the hydrogen atom that conceals the Runge–Lenz vector.

History

1925

— Heisenberg, Born, and Jordan formulate matrix mechanics, giving quantized energy levels a firm algebraic footing

1926

— Schrödinger's wave equation makes separable box and oscillator problems routinely solvable

1927

— Sommerfeld applies the free-electron model to metals, treating conduction electrons as a gas in a box and using the density of states to explain heat capacity and conductivity

1928

— Bloch extends the free-electron gas to periodic potentials, launching the band theory of solids

Going further

References. Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., §4.1 (the 3D box and separation of variables) · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §3 · Ashcroft & Mermin, Solid State Physics, Ch. 2 (the free-electron gas and density of states) · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. I, Complement on the infinite well. The low-level degeneracy table $g(3)=1$ , $g(6)=g(9)=g(11)=3$ , $g(12)=1$ , $g(14)=6$ follows by brute enumeration of ordered triples. The accidental degeneracy at $N=27$ is confirmed by noting that $(1,1,5)$ and $(3,3,3)$ are distinct sorted triples with $g(27)=4$ , and the level ordering is $N=3,6,9,11$ . The Weyl density-of-states asymptotic gives the number of states below radius $R$ as $(\pi/6)R^3-(3\pi/8)R^2$ , accurate to $0.02\%$ at $R=80$ . The numerical values $E_0\approx0.377\ \mathrm{eV}$ and $E_{111}\approx1.13\ \mathrm{eV}$ for a $1\ \mathrm{nm}$ electron box follow directly from $E_0=\hbar^2\pi^2/2m_eL^2$ .

Builds on: Infinite Square Well · Stationary States and Separation of Variables · Time-Independent Schrödinger Equation. Leads to: the density of states and free-electron model of solids, the isotropic harmonic oscillator, and degenerate perturbation theory.

How is the 3D box different from just solving the 1D well three times?

Mechanically it is three 1D wells, and that is the point of separation of variables: the wavefunction factorizes and the energies add. What is genuinely new is the combination of the three energies. In 1D the levels form a simple ladder with no degeneracy. In 3D the energy depends on the sum $n_1^2+n_2^2+n_3^2$ , so many different assignments of the excitation among the axes give the same total — degeneracy. So the individual factors are old, but the pattern of energies, and in particular the degeneracy structure, is a real three-dimensional phenomenon.

What exactly is "accidental" about the degeneracy at N = 27?

"Symmetry" degeneracy is the kind you can explain by an operator that maps one state to another — here, permuting the axes of the cube. All the states $(2,1,1),(1,2,1),(1,1,2)$ are related that way, so their equal energy is expected. But $(1,1,5)$ and $(3,3,3)$ both give $N=27$ while being different multisets of quantum numbers: no permutation of the cube's axes turns one into the other. Their coincidence is a property of the arithmetic of sums of three squares, not of any geometric symmetry of the box — hence "accidental." In some systems such accidents secretly reflect a hidden symmetry (the hydrogen atom is the famous case); in the cubic box the $N=27$ collision is a genuine number-theoretic coincidence.

Why does the degeneracy jump around so erratically as energy increases?

Because $g(N)$ counts integer lattice points on a sphere of radius $\sqrt N$ , which is an irregular, number-theoretic quantity. Some radii pass through many lattice points, others through few; there is no smooth formula. That is why you see $g=1,3,3,3,1,6,\dots$ with no obvious trend at low energy. Only when you average over many nearby levels — the large- $N$ limit — does the count smooth out into the density of states $\propto\sqrt E$ . The erratic microscopic degeneracy and the smooth macroscopic density of states are the same physics viewed at two scales.

Does a rectangular (non-cubic) box have any degeneracy?

Generically no — making all three side lengths different lifts the permutation degeneracy, because now $n_1^2/L_1^2+n_2^2/L_2^2+n_3^2/L_3^2$ changes when you permute the quantum numbers across unequal axes. You can still get accidental degeneracies if the ratios of the side lengths are special (for example commensurate), but for generic incommensurate sides every level is simple. This is the general lesson that degeneracy is tied to symmetry: remove the symmetry (equal sides) and the degeneracy it protected disappears.

How does the box lead to the free-electron model of a metal?

Take a macroscopic box, count how many single-particle states lie below a given energy (the states with $n_1^2+n_2^2+n_3^2\le R^2$ , which grows as the box volume), then fill those states with electrons two-at-a-time (spin up and down) up to the Fermi energy. The resulting density of states $g(E)\propto V\sqrt E$ is exactly the quantity Sommerfeld used in 1927 to compute the electronic heat capacity, magnetic susceptibility, and Fermi surface of metals. The box supplies the level counting; the Pauli principle supplies the filling. Remarkably, this crude model — free electrons in a box, ignoring the ionic lattice entirely — already gets the qualitative electronic physics of simple metals right.

Where does the factor √(8/L₁L₂L₃) in the wavefunction come from?

Each 1D factor $\sin(n_i\pi x/L_i)$ normalizes with a prefactor $\sqrt{2/L_i}$ , because $\int_0^{L_i}\sin^2(n_i\pi x/L_i)\,dx=L_i/2$ . Multiplying the three factors multiplies their normalization constants: $\sqrt{2/L_1}\cdot\sqrt{2/L_2}\cdot\sqrt{2/L_3}=\sqrt{8/L_1L_2L_3}$ . So the $8$ is just $2^3$ , one factor of $2$ per dimension. For a cube this is $\sqrt{8}/L^{3/2}=(2/L)^{3/2}$ .

Is there any physical meaning to the ground-state energy being 3E₀ rather than zero?

Yes — it is the three-dimensional zero-point energy of confinement. Even in its lowest state $(1,1,1)$ the particle cannot sit still: localizing it to a region of size $L$ forces, by the uncertainty principle, a minimum momentum spread and hence a minimum kinetic energy along each axis. The three axes each contribute $E_0=\hbar^2\pi^2/2mL^2$ , so the total floor is $3E_0$ . Shrinking the box raises this floor as $1/L^2$ , which is why confining electrons to nanometer dots produces electron-volt-scale energies and visible optical gaps.